Question

If the lines \(L_1:\frac{x-1}{-3} = \frac{y-2}{2k} = \frac{z-3}{2},L_2:\frac{x-1}{3k} = \frac{y-5}{1} = \frac{z-6}{-5}\) are perpendicular to each other, then the equation of a plane containing the line \(L_1\) and parallel to the line \(L_2\) for the value of \(k\) that satisfies this condition is ...

• A. \(31x+21y+46z-211 = 0\)
• B. \(602x-1155y-747z+3949 = 0\)
• C. \(43x+105y-154z+209 = 0\)
• D. \(2x-3y+5z-11 = 0\)

Hint:

Find $k$ from the perpendicular condition, then take the cross product of the directions as the normal.

Answer 1

The Correct Option is B

Step 1: Check by elimination
The normal must be perpendicular to both directions. Test option (B) with $(-21,-20,14)$: $602(-21)+1155\cdot20-747\cdot14=-12642+23100-10458=0$.
With $(-30,7,-35)$: $-18060-8085+26145=0$. And $(1,2,3)$: $602-2310-2241+3949=0$. Only (B) passes.

Final Answer:
Option (B). \[ \boxed{\text{(B)}} \]